Equação Cúbica

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The non-zero numbers a, b, c, etc., form a harmonic progression if their reciprocals 1/a, 1/b, 1/c, etc., form an arithmetic progression. Example The progression 1, s, 1 /5, 1 /7, ..., 1 /31 is harmonic since 1, 3, 5, 7, . . . , 31 form an arithmetic progression. The harmonic mean of two numbers a, b is 2ab/(a + b). PERMUTATIONS, COMBINATIONS, AND PROBABILITY Each separate arrangement of all or a part of a set of things is called a permutation. The number of permutations of n things taken r at a time, written P(n, r) = = n(n - 1)(n - 2) ⋅⋅⋅ (n - r + 1) Example The permutations of a, b, c two at a time are ab, ac, ba, ca, cb, and bc. The formula is P(3,2) = 3!/1! = 6. The permutations of a, b, c three at a time are abc, bac, cab, acb, bca, and cba. Each separate selection of objects that is possible irrespective of the order in which they are arranged is called a combination. The number of combinations of n things taken r at a time, written C(n, r) = n!/ [r!(n - r)!]. Example The combinations of a, b, c taken 2 at a time are ab, ac, bc; taken 3 at a time is abc. An important relation is r! C(n, r) = P(n, r). If an event can occur in p ways and fail to occur in q ways, all ways being equally likely, the probability of its occurrence is p/(p + q), and that of its failure q/(p + q). Example Two dice may be thrown in 36 separate ways. What is the prob- ability of throwing such that their sum is 7? Seven may arise in 6 ways: 1 and 6, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1. The probability of shooting 7 is j. THEORY OF EQUATIONS Linear Equations A linear equation is one of the first degree (i.e., only the first powers of the variables are involved), and the process of obtaining definite values for the unknown is called solving the equation. Every linear equation in one variable is written Ax + B = 0 or x =-B/A. Linear equations in n variables have the form a 11 x 1 + a 12 x 2 + ⋅⋅⋅ + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + ⋅⋅⋅ + a 2n x n = b 2 a m1 x 1 + a m2 x 2 + ⋅⋅⋅ + a mn x n = b m The solution of the system may then be found by elimination or matrix methods if a solution exists (see “Matrix Algebra and Matrix Compu- tations”). Quadratic Equations Every quadratic equation in one variable is expressible in the form ax 2 + bx + c = 0. a 0. This equation has two solutions, say, x 1 , x 2 , given by = If a, b, c are real, the discriminant b 2 - 4ac gives the character of the roots. If b 2 - 4ac > 0, the roots are real and unequal. If b 2 - 4ac < 0, the roots are complex conjugates. If b 2 - 4ac = 0 the roots are real and equal. Two quadratic equations in two variables can in general be solved only by numerical methods (see “Numerical Analysis and Approxi- mate Methods”). If one equation is of the first degree, the other of the second degree, a solution may be obtained by solving the first for one unknown. This result is substituted in the second equation and the resulting quadratic equation solved. Cubic Equations A cubic equation, in one variable, has the form x 3 + bx 2 + cx + d = 0. Every cubic equation having complex coefficients -b b 2 - 4 ac 2a x 1 x 2 n! (n - r)! 3-14 MATHEMATICS has three complex roots. If the coefficients are real numbers, then at least one of the roots must be real. The cubic equation x 3 + bx 2 + cx + d = 0 may be reduced by the substitution x = y - (b/3) to the form y 3 + py + q = 0, where p = s(3c - b 2 ), q = 1 / 27(27d - 9bc + 2b 3 ). This equa- tion has the solutions y 1 = A + B, y 2 =-a(A + B) + (i3 /2)(A - B), y 3 =-a(A + B) - (i3 /2)(A - B), where i 2 =-1, A = 3 - q /2 + R , B = 3 - q /2 - R , and R = (p/3) 3 + (q/2) 2 . If b, c, d are all real and if R > 0, there are one real root and two conjugate complex roots; if R = 0, there are three real roots, of which at least two are equal; if R < 0, there are three real unequal roots. If R < 0, these formulas are imprac- tical. In this case, the roots are given by y k = 2 - p /3 cos [(φ/3) + 120k], k = 0, 1, 2 where φ= cos -1 and the upper sign applies if q > 0, the lower if q < 0. Example x 3 + 3x 2 + 9x + 9 = 0 reduces to y 3 + 6y + 2 = 0 under x = y - 1. Here p = 6, q = 2, R = 9. Hence A = 3 2 , B = 3 - 4 . The desired roots in y are 3 2 - 3 - 4 and -a( 3 2 - 3 4 ) (i3 /2)( 3 2 + 3 4 ). The roots in x are x = y - 1. Example y 3 - 7y + 7 = 0. p =-7, q = 7, R < 0. Hence x k =- cos + 120k where φ= , = 3°3752. The roots are approximately -3.048916, 1.692020, and 1.356897. Example Many equations of state involve solving cubic equations for the compressibility factor Z. For example, the Redlich-Kwong-Soave equation of state requires solving Z 3 - Z 2 + cZ + d = 0, d < 0 where c and d depend on critical constants of the chemical species. In this case, only positive solutions, Z > 0, are desired. Quartic Equations See Ref. 118. General Polynomials of the nth Degree Denote the general polynomial equation of degree n by P(x) = a 0 x n + a 1 x n - 1 + ⋅⋅⋅ + a n - 1 x + a n = 0 If n > 4, there is no formula which gives the roots of the general equation. For fourth and higher order (even third order), the roots can be found numerically (see “Numerical Analysis and Approximate Methods”). However, there are some general theorems that may prove useful. Remainder Theorems When P(x) is a polynomial and P(x) is divided by x - a until a remainder independent of x is obtained, this remainder is equal to P(a). Example P(x) = 2x 4 - 3x 2 + 7x - 2 when divided by x + 1 (here a =-1) results in P(x) = (x + 1)(2x 3 - 2x 2 - x + 8) - 10 where -10 is the remainder. It is easy to see that P(-1) =-10. Factor Theorem If P(a) is zero, the polynomial P(x) has the fac- tor x - a. In other words, if a is a root of P(x) = 0, then x - a is a factor of P(x). If a number a is found to be a root of P(x) = 0, the division of P(x) by (x - a) leaves a polynomial of degree one less than that of the original equation, i.e., P(x) = Q(x)(x - a). Roots of Q(x) = 0 are clearly roots of P(x) = 0. Example P(x) = x 3 - 6x 2 + 11x - 6 = 0 has the root + 3. Then P(x) = (x - 3)(x 2 - 3x + 2). The roots of x 2 - 3x + 2 = 0 are 1 and 2. The roots of P(x) are therefore 1, 2, 3. Fundamental Theorem of Algebra Every polynomial of degree n has exactly n real or complex roots, counting multiplicities. φ 3 27 28 φ 3 28 3 q 2 /4 -p 3 /27

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Transcript of Equação Cúbica

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The non-zero numbers a, b, c, etc., form a harmonic progression iftheir reciprocals 1/a, 1/b, 1/c, etc., form an arithmetic progression.

Example The progression 1, s, 1⁄5, 1⁄7, . . . , 1⁄31 is harmonic since 1, 3, 5,7, . . . , 31 form an arithmetic progression.

The harmonic mean of two numbers a, b is 2ab/(a + b).

PERMUTATIONS, COMBINATIONS, AND PROBABILITY

Each separate arrangement of all or a part of a set of things is called apermutation. The number of permutations of n things taken r at atime, written

P(n, r) = = n(n − 1)(n − 2) ⋅⋅⋅ (n − r + 1)

Example The permutations of a, b, c two at a time are ab, ac, ba, ca, cb,and bc. The formula is P(3,2) = 3!/1! = 6. The permutations of a, b, c three at atime are abc, bac, cab, acb, bca, and cba.

Each separate selection of objects that is possible irrespective of theorder in which they are arranged is called a combination. The numberof combinations of n things taken r at a time, written C(n, r) = n!/[r!(n − r)!].

Example The combinations of a, b, c taken 2 at a time are ab, ac, bc; taken3 at a time is abc.

An important relation is r! C(n, r) = P(n, r).If an event can occur in p ways and fail to occur in q ways, all ways

being equally likely, the probability of its occurrence is p/(p + q), andthat of its failure q/(p + q).

Example Two dice may be thrown in 36 separate ways. What is the prob-ability of throwing such that their sum is 7? Seven may arise in 6 ways: 1 and 6,2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1. The probability of shooting 7 is j.

THEORY OF EQUATIONS

Linear Equations A linear equation is one of the first degree(i.e., only the first powers of the variables are involved), and theprocess of obtaining definite values for the unknown is called solvingthe equation. Every linear equation in one variable is written Ax + B =0 or x = −B/A. Linear equations in n variables have the form

a11x1 + a12 x2 + ⋅⋅⋅ + a1nxn = b1

a21x1 + a22 x2 + ⋅⋅⋅ + a2nxn = b2

:

am1x1 + am2x2 + ⋅⋅⋅ + amnxn = bm

The solution of the system may then be found by elimination or matrixmethods if a solution exists (see “Matrix Algebra and Matrix Compu-tations”).

Quadratic Equations Every quadratic equation in one variableis expressible in the form ax 2 + bx + c = 0. a ≠ 0. This equation has twosolutions, say, x1, x2, given by

6 =

If a, b, c are real, the discriminant b2 − 4ac gives the character of theroots. If b2 − 4ac > 0, the roots are real and unequal. If b2 − 4ac < 0, theroots are complex conjugates. If b2 − 4ac = 0 the roots are real andequal.

Two quadratic equations in two variables can in general be solvedonly by numerical methods (see “Numerical Analysis and Approxi-mate Methods”). If one equation is of the first degree, the other of thesecond degree, a solution may be obtained by solving the first for oneunknown. This result is substituted in the second equation and theresulting quadratic equation solved.

Cubic Equations A cubic equation, in one variable, has the formx3 + bx2 + cx + d = 0. Every cubic equation having complex coefficients

−b 6 Ïbw2w−w 4wacw}}

2a

x1

x2

n!}(n − r)!

3-14 MATHEMATICS

has three complex roots. If the coefficients are real numbers, then atleast one of the roots must be real. The cubic equation x3 + bx2 + cx +d = 0 may be reduced by the substitution x = y − (b/3) to the form y3 +py + q = 0, where p = s(3c − b2), q = 1⁄27(27d − 9bc + 2b3). This equa-tion has the solutions y1 = A + B, y2 = −a(A + B) + (iÏ3w/2)(A − B),

y3 = −a(A + B) − (iÏ3w/2)(A − B), where i2 = −1, A = Ï3

−wqw/2w +w ÏwRww,

B = Ï3

−wqw/2w −w ÏwRww, and R = (p/3)3 + (q/2)2. If b, c, d are all real and ifR > 0, there are one real root and two conjugate complex roots; if R =0, there are three real roots, of which at least two are equal; if R < 0,there are three real unequal roots. If R < 0, these formulas are imprac-tical. In this case, the roots are given by yk = 7 2 Ï−wpw/3w cos [(φ/3) +120k], k = 0, 1, 2 where

φ = cos−1 !§and the upper sign applies if q > 0, the lower if q < 0.

Example x3 + 3x2 + 9x + 9 = 0 reduces to y3 + 6y + 2 = 0 under x = y − 1.

Here p = 6, q = 2, R = 9. Hence A = Ï3

2w, B = Ï3

−w4w. The desired roots in y are

Ï3

2w − Ï3

−w4w and −a(Ï3

2w − Ï3

4w) 6 (iÏ3w/2)(Ï3

2w + Ï3

4w). The roots in x are x =

y − 1.

Example y3 − 7y + 7 = 0. p = −7, q = 7, R < 0. Hence

xk = −!§ cos 1 + 120k2where φ = !§, = 3°37′52″.

The roots are approximately −3.048916, 1.692020, and 1.356897.

Example Many equations of state involve solving cubic equations for thecompressibility factor Z. For example, the Redlich-Kwong-Soave equation ofstate requires solving

Z3 − Z2 + cZ + d = 0, d < 0

where c and d depend on critical constants of the chemical species. In this case,only positive solutions, Z > 0, are desired.

Quartic Equations See Ref. 118.General Polynomials of the nth Degree Denote the general

polynomial equation of degree n by

P(x) = a0 x n + a1x n − 1 + ⋅⋅⋅ + an − 1x + an = 0

If n > 4, there is no formula which gives the roots of the generalequation. For fourth and higher order (even third order), the rootscan be found numerically (see “Numerical Analysis and ApproximateMethods”). However, there are some general theorems that mayprove useful.

Remainder Theorems When P(x) is a polynomial and P(x) isdivided by x − a until a remainder independent of x is obtained, thisremainder is equal to P(a).

Example P(x) = 2x4 − 3x2 + 7x − 2 when divided by x + 1 (here a = −1)results in P(x) = (x + 1)(2x3 − 2x2 − x + 8) − 10 where −10 is the remainder. It iseasy to see that P(−1) = −10.

Factor Theorem If P(a) is zero, the polynomial P(x) has the fac-tor x − a. In other words, if a is a root of P(x) = 0, then x − a is a factorof P(x).

If a number a is found to be a root of P(x) = 0, the division of P(x) by(x − a) leaves a polynomial of degree one less than that of the originalequation, i.e., P(x) = Q(x)(x − a). Roots of Q(x) = 0 are clearly roots ofP(x) = 0.

Example P(x) = x3 − 6x2 + 11x − 6 = 0 has the root + 3. Then P(x) =(x − 3)(x2 − 3x + 2). The roots of x2 − 3x + 2 = 0 are 1 and 2. The roots of P(x) aretherefore 1, 2, 3.

Fundamental Theorem of Algebra Every polynomial of degreen has exactly n real or complex roots, counting multiplicities.

φ}3

27}28

φ}3

28}3

q2/4}−p3/27

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